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Last time, we went through the
same arguments that Maxwell and
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Boltzmann did to understand the
microscopic origin of the ideal
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gas law, PV equal nRT.
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And we saw how Maxwell had
hypothesized that the pressure
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of a gas on the walls of some
container, if the molecules were
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moving and they collided onto
the walls, that pressure must be
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due to the individual impacts of
the molecules on the wall.
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That must be due to the change
in the momentum of the wall when
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the molecules slammed right into
it.
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And that change in momentum
over some change in time is this
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force.
That force divided by the total
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area is the pressure.
And it was using that argument
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that we came up with an
expression, just like Maxwell
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did for the pressure times the
volume.
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And we see that we were able to
write it in terms of the
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velocity of the molecules that
were moving in this gas.
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Well, that is very nice because
if we have this theoretical
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expression P times V,
and we know experimentally that
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P times V equal nRT,
if this theory is
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correct, then this quantity n M,
average of the velocity
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squared, over three,
that better be equal to nRT.
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Or, solving for what we call
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the root mean square velocity,
that better be equal to the
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square root of 3RT over M.
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That kinetic theory made a
prediction for what the velocity
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ought to be.
It took, of course,
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another hundred years before
somebody could measure that.
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And, of course,
it is correct.
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But what is interesting here is
that this model gave us an
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understanding for what
temperature is.
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Temperature is a measure of the
speed of the molecules in the
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gas.
It is also, as we saw last
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time, a measure of their kinetic
energy.
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We saw that the kinetic energy
is one-half M average of the
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velocity squared. We put
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in our result from kinetic
theory.
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That shows us that the average
energy is three-halves RT.
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Temperature is the measure of
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the kinetic energy of these
molecules in the gas.
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For the first time,
there was a microscopic
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understanding of what
temperature was.
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Now, what we looked at also,
last time, was the
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Maxwell-Boltzmann distribution
of velocities or speeds and
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talked about this particular
functional form.
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How there was this quadratic
dependence right here,
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a low v because of the v
squared here and then an
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exponentially decaying tail.
There were two parameters,
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the mass and the temperature.
We talked about that last time.
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But, of course,
if there is a distribution here
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of speeds where this
distribution is a probability of
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finding a molecule with a
particular speed between v and v
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plus dv, --
-- then there also has to be a
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distribution of energies because
the velocity is related to the
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energy.
What do we have to do?
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We have to take that
Maxwell-Boltzmann speed
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distribution and change the
variable to energy.
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We know how to do that.
We know how to equate those two
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distribution functions.
And, using the Jacobean of the
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transformation that we talked
about last time,
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we can calculate what the
energy distribution is.
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f of E is our
Maxwell-Boltzmann energy
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distribution.
It is the probability of
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finding a molecule with a
kinetic energy E to E plus dE.
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It also has a decaying
exponential term,
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here, with the energy in the
argument, and it is multiplied
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by E of the one-half.
Let's take a look at what that
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distribution function looks
like.
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Here it is.
Notice, as we learned last
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time, that the energy has
nothing to do with the mass of
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the particle.
The only parameter that is
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important is the temperature.
At a given temperature,
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all particles,
it doesn't matter what their
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mass is, have the same energy.
So, all particles at
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degrees Kelvin have a
Maxwell-Boltzmann energy
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distribution that looks like
this.
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There is a very rapid rise in
that distribution function,
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unlike the velocity
distribution,
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which increased as v squared.
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It peaks rapidly,
and then there is an
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exponentially decaying term in
the energy.
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At 1500 degrees Kelvin,
the energy distribution looks
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like this.
The energy of the molecules has
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increased.
Since this is a probability,
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and the area under these curves
has to equal to one,
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if the energy goes up,
that is we have more molecules
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out here with higher energies,
well, then this maximum value
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for the probabilities has got to
go down because we have to keep
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the area under that curve equal
to one, since this is a
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probability.
What we see here,
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as we raise the energy,
is that we have more and more
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molecules with high energies.
There are not a lot of them,
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but there are some.
And they are going to be really
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important in the example I am
going to show you in just a
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moment.
Average energy at 600 Kelvin,
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just three-halves RT,
is 7.5 kilojoules per mole.
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Average energy at 1500 Kelvin
is 18.7 kilojoules per mole.
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But now, I want to show you an
example of the importance of
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this Maxwell-Boltzmann
distribution function for
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energies to chemical reactions,
the importance of it in making
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some kinds of chemical reactions
actually work.
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And so the example is going to
be this reaction.
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This reaction is called steam
reforming of natural gas.
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It is the reaction of methane
plus water.
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Natural gas is mostly methane.
But this reaction of methane
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and water makes CO and hydrogen.
It
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turns out that this reaction
does not work in the gas phase.
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That is, if you have a methane
molecule and water molecule
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collide, they are just going to
collide, bounce apart and go in
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their opposite directions.
They are not going to make CO
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and hydrogen.
And so what you have to have in
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this reaction to make it go is a
catalyst.
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That catalyst is going to be a
nickel surface.
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A catalyst is something that
will lower the activation energy
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barrier by changing the
mechanism of a reaction so that
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the reaction can proceed.
In this case,
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what happens is that the
methane and the water impinge on
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this nickel metal catalyst.
And the methane and the water
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decompose.
They fall apart to their
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elements on that nickel surface.
And then, once you have carbon,
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hydrogen and oxygen on that
nickel surface,
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the atoms rearrange and come
off as CO and molecular
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hydrogen.
So, that nickel surface is a
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catalyst.
We call this the catalytic
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reaction.
In particular,
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we call it a heterogeneous
catalytic reaction.
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It is heterogeneous because the
catalyst is in a different phase
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than the reactants.
The catalyst here is a solid.
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The reactants are gases.
A homogeneous catalyst is one
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in which the phase of the
catalyst and the reactants is
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the same.
This is a heterogeneous
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catalytic reaction.
It turns out that this reaction
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here is really an important
reaction from the standpoint of
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the production of hydrogen.
All of the hydrogen that you
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use, you know,
if you are doing a laboratory
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experiment and you go get a tank
of hydrogen, that hydrogen is
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made by this reaction,
by reacting methane and water
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to form that hydrogen.
All of our commercial sources
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of hydrogen come from carrying
out this reaction.
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For example,
if you are in the business of
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ammonia synthesis,
which is taking hydrogen and
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nitrogen to make ammonia,
which is also carried out on an
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iron surface,
another heterogeneous catalytic
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reaction.
And ammonia,
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of course, is the starting
material for lots of chemicals,
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in particular fertilizers.
If you have an ammonia plant,
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right next to the ammonia plant
you have a steamer-forming plant
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to make the hydrogen to feed
into this reaction.
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And another place where this is
useful is in methanol synthesis.
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That is, you can take hydrogen
and CO on a copper zinc oxide
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heterogeneous catalyst and make
methanol, the starting point for
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gasoline, which now is certainly
economically feasible.
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But it turns out that this
reaction here is really a hard
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one to carry out.
Despite the fact that nickel is
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a catalyst for the reaction and
makes it go, the reaction still
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has to be carried out at very
high temperatures.
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1500 degrees kelvin is a very
high temperature.
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It also needs very high
pressure.
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And one of the reasons why we
don't have a hydrogen economy is
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because it is so difficult to
make.
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Hydrogen can be done,
and it is done in all of the
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commercial processes that I
mentioned to you,
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but it is still hard.
1500 degrees kelvin,
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high pressures.
The question is,
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why is it so hard?
Why is there this barrier here?
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Why do we need to raise the
temperature of the gas in order
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to get this reaction to go?
Well, let's take a look at
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that.
What there exists here in this
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problem is what is called an
activation energy barrier to
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making the reaction go.
It turns out that this
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activation energy barrier is
really in the very first step of
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the reaction,
which is pulling the methane
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apart, pulling that first
hydrogen atom off of the methane
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molecule.
What I represent right here is
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the energy of the reaction as a
function of the reaction,
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and the reaction here is just
the first step.
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It is taking methane gas,
pulling the hydrogen off so
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that you have a methyl radical
stuck to the surface and you
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have a hydrogen atom stuck to
the surface.
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It is just breaking that first
C-H bond.
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What you can see here is that
you have to put 50 kilojoules of
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energy into that reaction in
order to make it go.
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You have to put that 50
kilocalories of energy in first,
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before you get any energy back.
You can see that this is
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exothermic.
But you have to put this energy
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in, in order to get that
reaction to go.
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Well, how do we know that?
Let me back up a minute.
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Here is where the
Maxwell-Boltzmann energy
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distribution is important.
What I did was I took those
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Boltzmann energy distributions
that I plotted for you at
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Kelvin and 1500 Kelvin and
turned them on their side.
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This is essentially equal to
the probability of finding a
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molecule at a particular energy
versus the energy.
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Here is the Maxwell-Boltzmann
distribution at 600 degrees
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Kelvin.
It is peaked to very low
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energies.
And, if you look here in this
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Maxwell-Boltzmann tail,
there are not very many
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molecules that have enough
energy to get over that barrier.
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However, if we raise the
temperature of the gas to
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degrees Kelvin,
then now you can see that there
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are more molecules here in the
high energy part of this tail.
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And it is these molecules at
these high energies that
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actually can get over this
barrier to dissociation of the
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methane molecule.
They are not many of those
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molecules with high energy,
but there are some.
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And that is important,
because what happens is when
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they have enough energy,
they react and they leave.
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And then this Boltzmann
distribution re-equilibrates.
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If the high energy molecules
leave, then there are lots of
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collisions here that kick some
more molecules up to the high
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energy part of the tail,
and they react.
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00:15:00 --> 00:15:04
And so it works.
That is why it is important.
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00:15:04 --> 00:15:09
How do we really know that
there is this barrier here to
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00:15:09 --> 00:15:15
the dissociation of methane,
and why is there this barrier?
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What is the physical origin of
this barrier to pulling this C-H
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bond apart?
Well, one way we could know
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that there was this barrier
there for sure is to do the
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following.
If we had a way to take methane
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gas and make it have just a
single energy,
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not the Maxwell-Boltzmann
distribution of energies that
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00:15:41 --> 00:15:46
have lots of different energies,
which have lots of different
225
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energies in them,
and you saw how broad those
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curves were.
But if we had a way to make
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methane gas with just say energy
E sub 1, whatever that is,
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we could then take those
molecules with just those
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energies, aim them at this
nickel surface,
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00:16:02 --> 00:16:06
and see if the methane fell
apart.
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00:16:06 --> 00:16:09
And, if it didn't,
then we would prepare molecules
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00:16:09 --> 00:16:13
at some higher energy and see if
they fell apart.
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00:16:13 --> 00:16:16
If they didn't then we would go
to higher energy.
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We would keep going until we
got to the top of the barrier.
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00:16:20 --> 00:16:24
Well, how do you do that?
How do you make molecules with
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a particular energy?
Because I just showed you a
237
00:16:28 --> 00:16:32
Maxwell-Boltzmann energy
distribution at 300 Kelvin,
238
00:16:32 --> 00:16:37
600 Kelvin.
There are a lot of energies
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present.
How do we make molecules with
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00:16:41 --> 00:16:46
just one energy?
We have a way to do that by
241
00:16:46 --> 00:16:52
using some beam techniques and
high pressure adiabatic
242
00:16:52 --> 00:16:57
expansions.
Basically, the way it works is
243
00:16:57 --> 00:17:01
this.
What we are going to do is take
244
00:17:01 --> 00:17:07
a tube here, which is going to
have a high pressure of methane
245
00:17:07 --> 00:17:10
in it.
Then we are going to punch a
246
00:17:10 --> 00:17:14
little hole in that tube.
This tube is actually sitting
247
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in a vacuum, so we are going to
expand the methane from that
248
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tube into the vacuum.
We are going to squirt it out.
249
00:17:24 --> 00:17:29
It is going to become a beam of
molecules.
250
00:17:29 --> 00:17:34
But when you expand a gas from
high pressure to low pressure,
251
00:17:34 --> 00:17:38
this expansion is called an
adiabatic expansion,
252
00:17:38 --> 00:17:42
which means the gas cools.
I will explain that in a
253
00:17:42 --> 00:17:47
moment, but the adiabatic
expansion you are going to talk
254
00:17:47 --> 00:17:50
a lot about in 5.60.
And, if you are a chemical
255
00:17:50 --> 00:17:55
engineer, you will talk even
more about it past 5.60.
256
00:17:55 --> 00:17:59
Here is how it works.
You take these molecules and
257
00:17:59 --> 00:18:04
squirt them out.
Because the pressure right here
258
00:18:04 --> 00:18:08
is so high, what happens is
there are lots and lots and lots
259
00:18:08 --> 00:18:11
of collisions.
And so, you can imagine,
260
00:18:11 --> 00:18:15
if you have some slow molecule
just kind of lumbering along and
261
00:18:15 --> 00:18:20
a fast one comes and hits you,
what is going to happen is you
262
00:18:20 --> 00:18:23
are going to speed up,
and the fast one is going to
263
00:18:23 --> 00:18:26
slow down.
And, if you keep doing that
264
00:18:26 --> 00:18:30
again and again and again,
all the molecules are going to
265
00:18:30 --> 00:18:35
end up with the same energy or
the same velocity.
266
00:18:35 --> 00:18:39
If you have so many collisions,
after a while they are going to
267
00:18:39 --> 00:18:44
all have the same velocity or
the same energy because one gets
268
00:18:44 --> 00:18:48
sped up, one gets slowed down.
They are all going to end up
269
00:18:48 --> 00:18:51
with the same energy.
That is what happens.
270
00:18:51 --> 00:18:56
That is how we make a beam of
molecules, a source of molecules
271
00:18:56 --> 00:19:00
with the same kinetic energy.
And basically,
272
00:19:00 --> 00:19:03
what is happening then is that
before the expansion,
273
00:19:03 --> 00:19:07
just look at this curve,
here is an actual Boltzmann
274
00:19:07 --> 00:19:10
distribution of velocities.
It is broad.
275
00:19:10 --> 00:19:14
But after this expansion,
here is the distribution.
276
00:19:14 --> 00:19:18
It is really pretty narrow.
If you put a temperature to
277
00:19:18 --> 00:19:21
that distribution,
you might find that at one
278
00:19:21 --> 00:19:24
degree Kelvin,
you can really make molecules
279
00:19:24 --> 00:19:30
with a single or very narrow
distribution in energies.
280
00:19:30 --> 00:19:33
That is what we do.
We make those molecules with
281
00:19:33 --> 00:19:38
this energy E sub 1.
Then we have some more tricks
282
00:19:38 --> 00:19:41
for changing those energies in a
controlled way.
283
00:19:41 --> 00:19:46
And we just keep cranking those
energies up and watch to see
284
00:19:46 --> 00:19:51
when the methane falls apart.
And so, you get data that kind
285
00:19:51 --> 00:19:54
of looks like this.
This is a dissociation
286
00:19:54 --> 00:19:58
probability of a methane as a
function of its energy.
287
00:19:58 --> 00:20:02
This is in kilocalories per
mole.
288
00:20:02 --> 00:20:05
At some point,
say right about here,
289
00:20:05 --> 00:20:10
15 kilocalories per mole,
the dissociation probability
290
00:20:10 --> 00:20:13
skyrockets.
All of a sudden,
291
00:20:13 --> 00:20:18
you are at a high enough energy
to get that methane to fall
292
00:20:18 --> 00:20:22
apart.
This tells you exactly where
293
00:20:22 --> 00:20:26
that barrier is.
But now, the question is,
294
00:20:26 --> 00:20:32
why is there this barrier?
Physically, why is there a
295
00:20:32 --> 00:20:38
barrier to pulling the hydrogen
off of the carbon when that
296
00:20:38 --> 00:20:43
methane molecule comes close to
the surface?
297
00:20:43 --> 00:20:48
Well, the reason is this.
In order to break the
298
00:20:48 --> 00:20:54
carbon-hydrogen bond in methane,
in order to have enough energy
299
00:20:54 --> 00:21:00
to break that bond,
what you need to do is you need
300
00:21:00 --> 00:21:05
to simultaneously form a
nickel-carbon bond and a
301
00:21:05 --> 00:21:10
nickel-hydrogen bond.
In other words,
302
00:21:10 --> 00:21:17
when this methane molecule here
comes into some nickel surface
303
00:21:17 --> 00:21:23
very slowly, because that
methane is tetrahedral and the
304
00:21:23 --> 00:21:30
carbon is kind of hidden in the
center, the hydrogens interact
305
00:21:30 --> 00:21:35
with the nickel.
But the carbon does not get in
306
00:21:35 --> 00:21:39
close enough to the nickel to
start to form a nickel-carbon
307
00:21:39 --> 00:21:42
bond.
Now, if you speed that methane
308
00:21:42 --> 00:21:46
molecule up and you really ram
it into the surface,
309
00:21:46 --> 00:21:49
upon collision of the molecule
with the surface,
310
00:21:49 --> 00:21:54
the hydrogens are pushed back,
the carbon gets in close enough
311
00:21:54 --> 00:22:00
to the nickel to start to form,
here, this nickel-carbon bond.
312
00:22:00 --> 00:22:05
You form a nickel-hydrogen
bond, and now you can break that
313
00:22:05 --> 00:22:09
C-H bond.
We call this black chemistry.
314
00:22:09 --> 00:22:12
The barrier,
there, to that reaction
315
00:22:12 --> 00:22:18
physically is the amount of
energy that you have to put into
316
00:22:18 --> 00:22:22
the molecule to push those
hydrogens back,
317
00:22:22 --> 00:22:30
to bend those hydrogens back to
distort this methane molecule.
318
00:22:30 --> 00:22:33
Once you do that,
the reaction goes.
319
00:22:33 --> 00:22:35
You are on a roll there.
It goes.
320
00:22:35 --> 00:22:39
So, that is the physical origin
of this barrier,
321
00:22:39 --> 00:22:43
here, to the dissociation of
methane.
322
00:22:43 --> 00:22:47
You need to put enough kinetic
energy into those methane
323
00:22:47 --> 00:22:51
molecules in order to get over
that barrier.
324
00:22:51 --> 00:22:56
And we do that on a practical
scale by raising the temperature
325
00:22:56 --> 00:23:01
of the gas.
And on a microscopic scale,
326
00:23:01 --> 00:23:06
this is what is happening.
You can imagine that my
327
00:23:06 --> 00:23:11
students and I did this
experiment 15 years ago.
328
00:23:11 --> 00:23:16
But then, we said if this
barrier here is the energy
329
00:23:16 --> 00:23:22
required to deform that methane
molecule, then we should be able
330
00:23:22 --> 00:23:26
to do this experiment.
This experiment is the
331
00:23:26 --> 00:23:31
following.
We are going to take our nickel
332
00:23:31 --> 00:23:34
surface, here,
and lower the temperature of
333
00:23:34 --> 00:23:38
the surface to 47 kelvin.
At that temperature,
334
00:23:38 --> 00:23:42
what we can do is we can freeze
a layer of methane on the
335
00:23:42 --> 00:23:44
surface.
We call it fizzy-sorb,
336
00:23:44 --> 00:23:47
a layer of methane on the
surface.
337
00:23:47 --> 00:23:51
It will just stick there.
And then, if this barrier is
338
00:23:51 --> 00:23:55
the energy required to deform or
to start the molecule,
339
00:23:55 --> 00:23:59
then in principle I could take
a hammer and pound that molecule
340
00:23:59 --> 00:24:04
into the correct shape for the
transition state that leads to
341
00:24:04 --> 00:24:09
dissociation.
It sounds like a simple idea,
342
00:24:09 --> 00:24:14
and it is, except that we
cannot really take that hammer.
343
00:24:14 --> 00:24:19
But we can take an argon atom.
We can freeze the methane onto
344
00:24:19 --> 00:24:23
the surface, and now we come in
with an argon atom.
345
00:24:23 --> 00:24:27
That is just a big ball.
Or, a xenon atom or a krypton
346
00:24:27 --> 00:24:30
atom.
And we know how to accelerate
347
00:24:30 --> 00:24:35
xenon or krypton.
We don't accelerate it too
348
00:24:35 --> 00:24:38
much, 50 kilocalories,
70 kilocalories per mole,
349
00:24:38 --> 00:24:42
something like that.
And then we can bring it in.
350
00:24:42 --> 00:24:47
What happens is that the impact
of the collision on that methane
351
00:24:47 --> 00:24:51
causes that methane molecule to
compress, distort,
352
00:24:51 --> 00:24:56
gets it into the configuration
of the transition state that
353
00:24:56 --> 00:25:00
leads to the methane falling
apart.
354
00:25:00 --> 00:25:02
And so, you get the same
result.
355
00:25:02 --> 00:25:05
The question,
there, is just getting the
356
00:25:05 --> 00:25:09
energy into the molecule to
actually distort it,
357
00:25:09 --> 00:25:14
to deform it so that you can
make the nickel-carbon bond and
358
00:25:14 --> 00:25:18
the nickel-hydrogen bond.
That is the key.
359
00:25:18 --> 00:25:22
My students and I,
after we had spent many years
360
00:25:22 --> 00:25:26
doing this experiment and spent
a lot of money on this
361
00:25:26 --> 00:25:31
experiment, said we have proven
for sure that when a bug flies
362
00:25:31 --> 00:25:36
into the windshield of your car,
you do get the same result as
363
00:25:36 --> 00:25:41
if you hit the bug on the
windshield of your car with a
364
00:25:41 --> 00:25:46
bug swatter.
That is what we get.
365
00:25:46 --> 00:25:51
This is an example of one
chemical reaction for which we
366
00:25:51 --> 00:25:55
know the physical origin of the
barrier.
367
00:25:55 --> 00:26:01
There are very few chemical
reactions for which we know
368
00:26:01 --> 00:26:06
anything about the physical
origin of a barrier to a
369
00:26:06 --> 00:26:11
reaction.
Now, what I want to talk about
370
00:26:11 --> 00:26:18
is go back to kinetic theory,
because the kinetic theory is
371
00:26:18 --> 00:26:25
going to allow us to calculate a
couple of more quantities that
372
00:26:25 --> 00:26:31
are of interest to us.
And one of those quantities is
373
00:26:31 --> 00:26:37
the collision frequency.
And let me just write that on
374
00:26:37 --> 00:26:41
the board here.
What I am going to want to
375
00:26:41 --> 00:26:47
calculate is something called
Z1, or I am going to call it Z1.
376
00:26:47 --> 00:26:53
It is the number of collisions
that a molecule makes in a gas
377
00:26:53 --> 00:26:58
per unit time.
And our unit time is going to
378
00:26:58 --> 00:27:06
be seconds.
This is the collision frequency
379
00:27:06 --> 00:27:13
of a single molecule in a gas.
380
00:27:13 --> 00:27:28
381
00:27:28 --> 00:27:29
How am I going to calculate the
quantity using the kinetic
382
00:27:29 --> 00:27:30
theory approach that we have
been talking about?
383
00:27:30 --> 00:27:33
What I am going to do is this.
I am going to take some gas,
384
00:27:33 --> 00:27:38
and the molecules in that gas
are represented by these blue
385
00:27:38 --> 00:27:42
circles right here.
They have a diameter D.
386
00:27:42 --> 00:27:47
And then, within that gas,
I am going to imagine this
387
00:27:47 --> 00:27:52
lighter blue cylinder.
That cylinder is an imaginary
388
00:27:52 --> 00:27:55
construct.
We are going to use it to
389
00:27:55 --> 00:28:00
calculate this collision
frequency.
390
00:28:00 --> 00:28:04
What it is going to be is the
collision volume.
391
00:28:04 --> 00:28:09
I have set the diameter of that
cylinder to be equal to two
392
00:28:09 --> 00:28:12
times the diameter of the
molecule.
393
00:28:12 --> 00:28:17
I have set this upright.
And the bottom line is that all
394
00:28:17 --> 00:28:22
of the molecules that
instantaneously happen to be in
395
00:28:22 --> 00:28:26
this cylinder when a molecule
comes through,
396
00:28:26 --> 00:28:32
all of those molecules are
actually going to be hit.
397
00:28:32 --> 00:28:37
They are going to suffer a
collision with another molecule.
398
00:28:37 --> 00:28:42
And I am going to use that,
the number of molecules in this
399
00:28:42 --> 00:28:46
volume, to calculate this
collision frequency.
400
00:28:46 --> 00:28:51
That is what I am going to do.
Snapshot in time,
401
00:28:51 --> 00:28:56
those molecules are frozen
there, except there is one
402
00:28:56 --> 00:29:02
smart-alecky molecule that comes
cruising on through.
403
00:29:02 --> 00:29:04
Here he comes.
Bam, bam, bam.
404
00:29:04 --> 00:29:10
Hits these three molecules,
here, because they are more
405
00:29:10 --> 00:29:14
than halfway into that collision
cylinder.
406
00:29:14 --> 00:29:20
What I know is this smart-aleck
was moving at some average
407
00:29:20 --> 00:29:24
velocity, which I am going to
call vbar.
408
00:29:24 --> 00:29:27
In a time T,
this molecule,
409
00:29:27 --> 00:29:32
since it is moving with an
average velocity vbar,
410
00:29:32 --> 00:29:36
in a time t,
it is traveling a length vbar
411
00:29:36 --> 00:29:41
times t.
412
00:29:41 --> 00:29:44
Velocity times time,
that is going to give you a
413
00:29:44 --> 00:29:47
length.
And so, I am going to set,
414
00:29:47 --> 00:29:50
for convenience,
this time equal to one second
415
00:29:50 --> 00:29:53
because that is going to be my
unit of time.
416
00:29:53 --> 00:29:58
I want to do this per second.
I am going to set this equal to
417
00:29:58 --> 00:30:02
one second.
Therefore, this molecule is
418
00:30:02 --> 00:30:07
going to travel a length vbar in
one second.
419
00:30:07 --> 00:30:13
And that is the length that I
am going to make the collision
420
00:30:13 --> 00:30:16
cylinder.
I am going to make that
421
00:30:16 --> 00:30:22
collision cylinder be vbar long.
That is the distance traveled
422
00:30:22 --> 00:30:28
in one second by this
smart-alecky molecule.
423
00:30:28 --> 00:30:34
Now, all I have to do is I have
got to take the volume of the
424
00:30:34 --> 00:30:40
collision cylinder and multiply
it by all the molecules that
425
00:30:40 --> 00:30:45
happen to be at the volume at
that particular time.
426
00:30:45 --> 00:30:51
Because that volume represents
the volume swept out in one
427
00:30:51 --> 00:30:55
second by this smart-alecky
molecule.
428
00:30:55 --> 00:31:00
What is the volume of the
cylinder?
429
00:31:00 --> 00:31:03
I set the diameter to be equal
to 2d.
430
00:31:03 --> 00:31:08
The volume of the cylinder is
the cross-sectional area times
431
00:31:08 --> 00:31:11
the length.
That cross-sectional area,
432
00:31:11 --> 00:31:16
then, is pi r squared,
but r here is equal to d,
433
00:31:16 --> 00:31:21
just to make it confusing.
The area is pi d squared.
434
00:31:21 --> 00:31:24
The length is vbar.
435
00:31:24 --> 00:31:28
That is the volume of the
cylinder, pi d squared vbar.
436
00:31:28 --> 00:31:34
So, we have the volume of the
437
00:31:34 --> 00:31:37
cylinder.
Now what we are going to need
438
00:31:37 --> 00:31:42
to know is the density of the
molecules in this cylinder,
439
00:31:42 --> 00:31:46
which is the density of the
molecules in the gas.
440
00:31:46 --> 00:31:50
The cylinder is not special.
The cylinder is an imaginary
441
00:31:50 --> 00:31:55
construct that I put in there to
help me calculate the collision
442
00:31:55 --> 00:31:58
frequency.
I need the density of the
443
00:31:58 --> 00:32:03
molecules in the gas or in the
cylinder.
444
00:32:03 --> 00:32:07
The density of the molecules I
am going to set as equal to N,
445
00:32:07 --> 00:32:10
the total number of molecules
in this volume,
446
00:32:10 --> 00:32:13
divided by the volume of the
gas.
447
00:32:13 --> 00:32:17
I am going to call that rho.
N over V is going to be equal
448
00:32:17 --> 00:32:21
to rho.
That is the number of molecules
449
00:32:21 --> 00:32:25
per cubic meter.
That is the density of the gas.
450
00:32:25 --> 00:32:29
This is the symbol that I am
going to use from now on for
451
00:32:29 --> 00:32:34
density defined as molecules per
cubic meter.
452
00:32:34 --> 00:32:39
Then the collision frequency is
simply going to be the density
453
00:32:39 --> 00:32:43
of the molecules times the
volume of the cylinder.
454
00:32:43 --> 00:32:47
The collision frequency Z1,
here, is this volume of the
455
00:32:47 --> 00:32:52
cylinder, pi d squared vbar
times the density.
456
00:32:52 --> 00:32:53
Why?
457
00:32:53 --> 00:32:58
Because I set it up so that
every molecule in that cylinder
458
00:32:58 --> 00:33:04
would suffer a collision.
And that cylinder is the length
459
00:33:04 --> 00:33:08
that a molecule travels in one
second.
460
00:33:08 --> 00:33:11
There is the density times the
volume.
461
00:33:11 --> 00:33:16
The meters cubed disappears.
What I have left is rho,
462
00:33:16 --> 00:33:20
the density,
pi d squared vbar.
463
00:33:20 --> 00:33:24
This is collisions per second.
464
00:33:24 --> 00:33:27
So, I have my collision
frequency.
465
00:33:27 --> 00:33:33
This is the single-molecule
collision frequency.
466
00:33:33 --> 00:33:37
This is the frequency of
collisions that one molecule
467
00:33:37 --> 00:33:40
makes with the other gas
molecules.
468
00:33:40 --> 00:33:44
That is important.
We are going to do a different
469
00:33:44 --> 00:33:47
kind of collision frequency in
just a moment.
470
00:33:47 --> 00:33:52
But now, it turns out that
there is another factor here
471
00:33:52 --> 00:33:57
that I had left out because I
haven't done as sophisticated of
472
00:33:57 --> 00:34:02
an analysis as I could.
And that is that I made the
473
00:34:02 --> 00:34:07
assumption in my picture before
that all the other molecules
474
00:34:07 --> 00:34:11
were frozen in time and that
there was only one smart-aleck
475
00:34:11 --> 00:34:15
that was moving around,
cruising through.
476
00:34:15 --> 00:34:17
The reality is they are all
moving.
477
00:34:17 --> 00:34:23
And what I really have to do is
I have to take into account the
478
00:34:23 --> 00:34:27
relative velocities of the
molecules, the relative speeds.
479
00:34:27 --> 00:34:31
I can do that.
There is a little more
480
00:34:31 --> 00:34:36
sophisticated analysis in doing
that, but I could do that.
481
00:34:36 --> 00:34:40
And, if we do that,
this makes a difference of the
482
00:34:40 --> 00:34:44
square root of two.
I am just going to put that
483
00:34:44 --> 00:34:49
square root of two in there
right now, and later on you will
484
00:34:49 --> 00:34:54
be able to see where that comes
from, in a later course.
485
00:34:54 --> 00:34:59
That actually is the collision
frequency of a single molecule
486
00:34:59 --> 00:35:01
in the gas.
487
00:35:01 --> 00:35:07
488
00:35:07 --> 00:35:12
I am also interested in another
quantity, which is the total
489
00:35:12 --> 00:35:16
collision frequency.
That was the single collision
490
00:35:16 --> 00:35:21
frequency, but I am also
interested in knowing how many
491
00:35:21 --> 00:35:26
collisions are occurring in the
entire gas per unit time,
492
00:35:26 --> 00:35:32
the total collision frequency.
You know why I am interested in
493
00:35:32 --> 00:35:36
that number?
I am interested in that number
494
00:35:36 --> 00:35:41
because that is going to be the
upper limit to any reaction
495
00:35:41 --> 00:35:44
rate.
A reaction in the gas phase or
496
00:35:44 --> 00:35:49
in solution cannot happen any
faster than the molecules
497
00:35:49 --> 00:35:52
collide.
They have to collide before a
498
00:35:52 --> 00:35:57
reaction is going to occur.
The total collision frequency,
499
00:35:57 --> 00:36:03
the importance of that number
is that it is the upper limit to
500
00:36:03 --> 00:36:08
a reaction rate.
Let's calculate that.
501
00:36:08 --> 00:36:13
Z is going to be equal to the
collision frequency of one
502
00:36:13 --> 00:36:17
molecule, Z1,
times the total number of
503
00:36:17 --> 00:36:21
molecules in the gas, N.
504
00:36:21 --> 00:36:25
That is what N stands for.
But since each collision
505
00:36:25 --> 00:36:31
involves two molecules,
I am going to have to multiply
506
00:36:31 --> 00:36:36
this by one-half.
Otherwise, I am going to over
507
00:36:36 --> 00:36:41
count the number of collisions
because each collision involves
508
00:36:41 --> 00:36:44
two molecules.
The total collision frequency
509
00:36:44 --> 00:36:48
here is one-half times N times
Z1.
510
00:36:48 --> 00:36:52
Therefore, if I go and plug in
my expression for Z1 and N and
511
00:36:52 --> 00:36:56
simplify things,
my total collision frequency
512
00:36:56 --> 00:37:01
here is one over the square root
of two times N times rho pi d
513
00:37:01 --> 00:37:06
squared times the average speed.
514
00:37:06 --> 00:37:11
That is my total collisions per
515
00:37:11 --> 00:37:15
second.
This is not rocket science.
516
00:37:15 --> 00:37:20
This is easy.
What I want you to realize is
517
00:37:20 --> 00:37:26
that you do have to understand
what N is, what rho is,
518
00:37:26 --> 00:37:32
what d is, what vbar is.
And I am always surprised on an
519
00:37:32 --> 00:37:38
exam, when we are going to give
you this equation,
520
00:37:38 --> 00:37:43
that students don't actually
know what rho is,
521
00:37:43 --> 00:37:48
or N is, or v is,
or d is.
522
00:37:48 --> 00:37:51
This is easy.
You do just have to understand,
523
00:37:51 --> 00:37:54
this is the total number of
molecules in the gas,
524
00:37:54 --> 00:37:58
this is the density of the gas
in molecules per cubic meter,
525
00:37:58 --> 00:38:02
the diameter of the molecule,
the average speed.
526
00:38:02 --> 00:38:07
That was just a helpful hint.
This is the upper limit to the
527
00:38:07 --> 00:38:10
reaction rate.
When somebody tells you,
528
00:38:10 --> 00:38:15
the rate of this reaction is so
and so many molecules per cubic
529
00:38:15 --> 00:38:18
meter per second,
what you can do is a
530
00:38:18 --> 00:38:24
back-of-the-envelope calculation
to see whether or not they are
531
00:38:24 --> 00:38:30
telling you the reaction rate is
greater than this number.
532
00:38:30 --> 00:38:35
If it is, you can say ah-ha,
got you, it cannot possibly be.
533
00:38:35 --> 00:38:40
Really important.
If a reaction has this rate,
534
00:38:40 --> 00:38:44
it will mean that the
probability of the reaction
535
00:38:44 --> 00:38:49
occurring is one.
If the reaction has that rate,
536
00:38:49 --> 00:38:52
we call that the gas kinetic
rate.
537
00:38:52 --> 00:38:55
That is a term that we also
use.
538
00:38:55 --> 00:38:59
Then, finally,
one other quantity from the gas
539
00:38:59 --> 00:39:03
kinetic theory.
Yes?
540
00:39:03 --> 00:39:10
541
00:39:10 --> 00:39:13
She asked, what velocity,
here, would you use?
542
00:39:13 --> 00:39:17
You actually are going to,
in that particular case,
543
00:39:17 --> 00:39:21
have to take the average of the
average velocities.
544
00:39:21 --> 00:39:24
In other words,
if you had two different
545
00:39:24 --> 00:39:29
molecules that were reacting for
this average velocity here,
546
00:39:29 --> 00:39:33
you are going to have to use
the average of the average
547
00:39:33 --> 00:39:38
velocity.
And then your error bars on the
548
00:39:38 --> 00:39:44
experiment will always be large
enough to take that into
549
00:39:44 --> 00:39:47
consideration.
Good question.
550
00:39:47 --> 00:39:52
One other quantity,
something called the mean free
551
00:39:52 --> 00:39:55
path.
What I want to know,
552
00:39:55 --> 00:40:00
here, is on the average,
how far does the molecule
553
00:40:00 --> 00:40:07
travel in the gas before it
suffers a collision?
554
00:40:07 --> 00:40:09
That is what a mean free path
is.
555
00:40:09 --> 00:40:13
A mean free path in solution.
Sometimes you will do a solid
556
00:40:13 --> 00:40:17
state physics course and will
hear about the electron mean
557
00:40:17 --> 00:40:20
free path.
A mean free path is always the
558
00:40:20 --> 00:40:22
distance traveled between
collisions.
559
00:40:22 --> 00:40:26
That is what that is.
And we are going to call it
560
00:40:26 --> 00:40:30
lambda.
This is not wavelength anymore.
561
00:40:30 --> 00:40:33
We just changed our definition
of lambda.
562
00:40:33 --> 00:40:36
Lambda, here,
is the average distance between
563
00:40:36 --> 00:40:40
collisions.
How are we going to calculate
564
00:40:40 --> 00:40:43
that?
What we are going to do is take
565
00:40:43 --> 00:40:47
the average distance that a
molecule travels per unit time,
566
00:40:47 --> 00:40:52
per second, and are going to
divide it by the number of
567
00:40:52 --> 00:40:55
collisions that occur per
second.
568
00:40:55 --> 00:40:59
And you are going to see that
the per seconds are going to
569
00:40:59 --> 00:41:03
cancel here.
We are going to have the
570
00:41:03 --> 00:41:08
average distance per collision.
That is what we are after.
571
00:41:08 --> 00:41:13
This is easy to do because the
average distance traveled per
572
00:41:13 --> 00:41:16
second is simply the average
velocity.
573
00:41:16 --> 00:41:20
It is meters per second.
The average distance traveled
574
00:41:20 --> 00:41:24
per second.
The number of collisions per
575
00:41:24 --> 00:41:27
second that happened,
we just calculated that.
576
00:41:27 --> 00:41:33
That was Z1.
We can substitute in Z1 vbar,
577
00:41:33 --> 00:41:38
vbar cancels,
and what we are left with is
578
00:41:38 --> 00:41:46
one over the square root of 2
rho pi d squared.
579
00:41:46 --> 00:41:49
That is meters.
580
00:41:49 --> 00:41:56
That is the average distance
the molecule traveled per
581
00:41:56 --> 00:42:01
collision.
This is another general generic
582
00:42:01 --> 00:42:06
quantity useful in many cases,
also in other kinds of cases
583
00:42:06 --> 00:42:10
that I just mentioned in solid
state physics.
584
00:42:10 --> 00:42:16
What I want to do now is that
we said we were going to start
585
00:42:16 --> 00:42:19
talking about the motions of
molecules.
586
00:42:19 --> 00:42:24
We have taken care of,
now, the translational motion.
587
00:42:24 --> 00:42:29
Now I want to start talking
about the internal degrees of
588
00:42:29 --> 00:42:33
freedom.
And let's do that.
589
00:42:33 --> 00:42:38
Let me just see if I have
something, here.
590
00:42:38 --> 00:42:43
Let me raise the board,
here.
591
00:42:43 --> 00:42:58
592
00:42:58 --> 00:43:02
We have talked about this
molecule, which is a set of
593
00:43:02 --> 00:43:07
atoms that are bonded together.
And, as we always say,
594
00:43:07 --> 00:43:12
they are bonded together for
the mutual comfort of their
595
00:43:12 --> 00:43:15
electrons.
And we saw that it is not
596
00:43:15 --> 00:43:17
frozen.
It moves.
597
00:43:17 --> 00:43:20
It has kinetic energy.
It has velocity.
598
00:43:20 --> 00:43:25
The velocity generates this
macroscopic property of
599
00:43:25 --> 00:43:30
pressure.
But molecules also jiggle.
600
00:43:30 --> 00:43:33
For example,
we are going to take here
601
00:43:33 --> 00:43:39
nitrogen, triple bond.
That triple bond actually
602
00:43:39 --> 00:43:44
functions like a spring.
We can think of that bond as a
603
00:43:44 --> 00:43:49
spring between the two atoms.
That spring can stretch,
604
00:43:49 --> 00:43:54
and that spring can compress.
When we tell you the
605
00:43:54 --> 00:44:00
equilibrium bond length of some
molecule is some number,
606
00:44:00 --> 00:44:05
it is the equilibrium bond
length.
607
00:44:05 --> 00:44:09
It is not the bond length when
the molecule is stretched.
608
00:44:09 --> 00:44:14
It is not the bond length when
the molecule is compressed.
609
00:44:14 --> 00:44:19
This motion is the vibrational
motion of the molecule.
610
00:44:19 --> 00:44:24
611
00:44:24 --> 00:44:30
So, we have that kind of
internal motion in a molecule.
612
00:44:30 --> 00:44:33
Molecules also tumble.
For example,
613
00:44:33 --> 00:44:38
let's take our nitrogen
molecule, again.
614
00:44:38 --> 00:44:45
It rotates around an axis.
Going through the center of the
615
00:44:45 --> 00:44:49
mass, it rotates around this
axis.
616
00:44:49 --> 00:44:54
This nitrogen molecule also
rotates around an axis,
617
00:44:54 --> 00:45:00
here, perpendicular to the
board.
618
00:45:00 --> 00:45:06
It rotates in that direction.
This is the rotational motion
619
00:45:06 --> 00:45:11
of the molecule.
The different ways in which we
620
00:45:11 --> 00:45:15
can have this motion are called
modes.
621
00:45:15 --> 00:45:21
This is a rotational mode.
This is vibrational mode.
622
00:45:21 --> 00:45:28
We did not use the word before
in talking about translation,
623
00:45:28 --> 00:45:33
but those are translational
modes.
624
00:45:33 --> 00:45:40
Sometimes, we also call these
modes degrees of freedom.
625
00:45:40 --> 00:45:49
A molecule has translational
modes or degrees of freedom,
626
00:45:49 --> 00:45:57
vibrational degrees of freedom,
and rotational degrees of
627
00:45:57 --> 00:46:00
freedom.
Now, in general,
628
00:46:00 --> 00:46:10
if you have an N-atom molecule,
you have 3N total modes.
629
00:46:10 --> 00:46:17
When I say total modes that
means the sum of translation,
630
00:46:17 --> 00:46:24
rotation and vibration.
Of those 3N total modes,
631
00:46:24 --> 00:46:30
three of them are translational
modes.
632
00:46:30 --> 00:46:35
633
00:46:35 --> 00:46:38
Why do we have three
translational modes?
634
00:46:38 --> 00:46:43
We have three translational
modes because we are operating
635
00:46:43 --> 00:46:48
in a three-dimensional space.
If we have a molecule here,
636
00:46:48 --> 00:46:53
that molecule has motion in the
x direction, in the y direction,
637
00:46:53 --> 00:46:57
and also in the z direction.
Those are the three
638
00:46:57 --> 00:47:02
translational modes of the
molecule.
639
00:47:02 --> 00:47:08
What that means then,
if three of the modes are used
640
00:47:08 --> 00:47:13
up for translation,
then we must have 3N minus 3
641
00:47:13 --> 00:47:20
internal modes.
3N minus 3 must be the number
642
00:47:20 --> 00:47:27
of modes that is leftover for
the internal degrees of freedom
643
00:47:27 --> 00:47:33
for rotation and also for
vibration.
644
00:47:33 --> 00:47:39
And what we are going to see,
next time, is that these
645
00:47:39 --> 00:47:46
internal degrees of freedom,
these internal modes are going
646
00:47:46 --> 00:47:51
to be quantized.
That is, this molecule is going
647
00:47:51 --> 00:47:59
to be able to vibrate with only
certain amounts of energy.
648
00:47:59 --> 00:48:02
There is going to be a ground
vibrational state.
649
00:48:02 --> 00:48:07
There is going to be a first
excited vibrational state.
650
00:48:07 --> 00:48:12
A second excited vibrational
state, just like we talked about
651
00:48:12 --> 00:48:17
the energy levels of a hydrogen
atom, where the electron was
652
00:48:17 --> 00:48:22
bound with different amounts of
energy dictated by a principle
653
00:48:22 --> 00:48:26
quantum number.
The vibrational degree of
654
00:48:26 --> 00:48:30
freedom is also going to be
dictated by a vibrational
655
00:48:30 --> 00:48:33
quantum number.
In addition,
656
00:48:33 --> 00:48:38
the rotations of the molecules,
they are also quantized.
657
00:48:38 --> 00:48:43
A molecule is going to be able
to rotate with only one energy,
658
00:48:43 --> 00:48:46
or another energy,
or another energy,
659
00:48:46 --> 00:48:49
but not any energies in
between.
660
00:48:49 --> 00:48:52
There are discrete rotational
states.
661
00:48:52 --> 00:48:56
We are going to be talking
about another kind of quantum
662
00:48:56 --> 00:48:58
number.
In that case,
663
00:48:58 --> 00:49:04
a rotational quantum number.
That is where we are going to
664
00:49:04.451 --> 49:07
pick up on Monday.
See you then.